# If a,b,c,d are positive integers with a sum of 63, what is the maximum value of ab + bc + cd ? (By using calculus to solve it)) [closed]

If $$a,b,c,d$$ are positive integers with $$a+b+c+d=63,$$ what is the maximum value of $$ab+bc+cd?$$

I treated this as an optimization problem and tried to solve it but I did not make any progress.

## closed as off-topic by José Carlos Santos, Shailesh, YuiTo Cheng, Jyrki Lahtonen, LeucippusJun 6 at 4:07

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• How did you try to solve this? Various methods are possible. You could usefully explain the context in which calculus arises - is this an example to illustrate a particular method? – Mark Bennet Jun 5 at 7:04

This was recently posted on youtube.

Anwyway, $$ab + bc + cd = (a+c)(b+d) - ad$$

To maximize the objective, set $$ad = 1$$ and make $$(a+c)\times (b+d)$$ as close to square as possible.

$$32\times 31 - 1 = 991$$

• I know. I just wanted to know how to solve this using calculus – Percy04 Jun 5 at 7:38
• Calculus applies to continuous functions. Restricting $a,b,c,d$ to integers suggests to me to try other avenues first. Calculus works best over real numbers. – Doug M Jun 5 at 7:45

Given $$a+b+c+d=63$$.

Denote $$f(a,b,c,d)=ab+bc+cd$$.

We solve the question by Lagranges multiplier method.

Consider $$F(a,b,c,d,\lambda)=ab+bc+cd+\lambda(a+b+c+d-63)$$, where $$\lambda$$ is Lagrange multiplier.

Applying necessary conditions for maximum of $$f$$ i.e $$\frac{∂F}{∂a}=\frac{∂F}{∂b}=\frac{∂F}{∂c}=\frac{∂F}{∂d}=\frac{∂F}{∂\lambda}=0$$

We get $$b+\lambda=0, a+c+\lambda=0, b+d+\lambda=0, c+\lambda=0\quad \text{and}\quad a+b+c+d=63$$

$$\implies a=0, d=0, \lambda=-\frac{63}{2}, c=\frac{63}{2}, b=\frac{63}{2}$$

Since we need $$a, b, c, d$$ to be integers, so for $$b$$ and $$c$$ we can test the closest values of $$31$$ and $$32$$.

Hence $$ab+bc+cd \lt 992$$.

Therefore maximum value of $$ab+bc+cd$$ is $$991$$.

Another way to see this:

Note that the function can be written $$ab+c(b+d)$$ with $$b+d\gt b$$ so given fixed $$b$$ and $$d$$ we want $$c$$ as large as possible and $$a$$ as small as possible, so $$a=1$$.

Similarly we can express the function as $$(a+c)b+cd$$ so $$d=1$$ and $$b$$ is as large as possible.

This gives us $$b+c+bc$$ as the total with $$b+c=61$$ so the maximum value of the function is the maximum of $$61+bc$$ for $$b+c=61$$

Then maximising a product when the sum is fixed is well known; the identity $$4bc=(b+c)^2-(b-c)^2$$ will do the trick - $$b$$ and $$c$$ have to be as close together as possible.

Let $$a=x+1,$$ $$b=y+1$$, $$c=z+1$$ and $$d=t+1$$.

Thus, $$x+y+z+t=59$$, where $$x,$$ $$y$$, $$z$$ and $$t$$ are non-negatives and by AM-GM we obtain: $$ab+bc+cd=(x+1)(y+1)+(y+1)(z+1)+(z+1)(t+1)=$$ $$=xy+yz+zt+x+2y+2z+t+3=xy+yz+zt+y+z+62\leq$$ $$\leq xy+yz+zt+tx+59+62=(x+z)(y+t)+121\leq$$ $$\leq\left(\frac{x+z+y+t}{2}\right)^2+121=991.25.$$ Id est, $$ab+bc+cd\leq991,$$ but for $$(a,b,c,d)=(1,30,31,1)$$ we have equality, which says that we got a maximal value.

• why did you replace y+z with 59+tx? Shouldn't it have been 59-t-x? – MathDude3013 Jun 17 at 14:40
• Because y+z<=59 – Michael Rozenberg Jun 17 at 15:30
• Dude, how did you get the 'tx' term? – MathDude3013 Jun 17 at 16:23
• @MathDude I just added this term and the sum was increased. – Michael Rozenberg Jun 17 at 16:25